JEE Main · 2019 · Shift-IeasyBOND-090

The type of hybridization and number of lone pair(s) of electrons of Xe in XeOF4, respectively, are:

Chemical Bonding · Class 11 · JEE Main Previous Year Question

Question

The type of hybridization and number of lone pair(s) of electrons of Xe in XeOF4\mathrm{XeOF_4}, respectively, are:

Options
  1. a

    sp3d2\mathrm{sp^3d^2} and 1

  2. b

    sp3d2\mathrm{sp^3d^2} and 2

  3. c

    sp3d\mathrm{sp^3d} and 2

  4. d

    sp3d\mathrm{sp^3d} and 1

Correct Answera

sp3d2\mathrm{sp^3d^2} and 1

Detailed Solution

🧠 Determine the hybridisation and lone pairs on Xe in \ceXeOF4\ce{XeOF4}.

Xe has 8 valence electrons. In \ceXeOF4\ce{XeOF4}: 4 bonds to F + 1 bond to O = 5 bonds total. LP on Xe = (8 − 5)/2 = 1 lone pair.

Steric number = 5 + 1 = 6sp³d² hybridisation → octahedral electron geometry → with 1 LP in one axial position, the molecular shape is square pyramidal.

⚡ Xe uses 3d orbitals to expand beyond the octet: 5 bonds + 1 LP = 12 electrons around Xe, requiring sp³d² hybridisation.

Answer: (a) sp3d2 and 1\boxed{\text{Answer: (a) sp}^3\text{d}^2\text{ and 1}}

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